Given ,
$x + y + z = 3$
$\Rightarrow x +\left(\frac{y}{2}\right) +\left(\frac{y}{2}\right) +\left(\frac{z}{3}\right) +\left(\frac{z}{3}\right) +\left(\frac{z}{3}\right) = 3.$
Now using A.M. G.M inequality , we have :
$\dfrac{\left[x +\left(\frac{y}{2}\right) +\left(\frac{y}{2}\right) +\left(\frac{z}{3}\right) +\left(\frac{z}{3}\right) +\left(\frac{z}{3}\right)\right]}{6}$
$\geq \large\left[x .\left(\frac{y}{2}\right) .\left(\frac{y}{2}\right) .\left(\frac{z}{3}\right) .\left(\frac{z}{3}\right) .\left(\frac{z}{3}\right)\right]^{\frac{1}{6}}$
$\Rightarrow \large\left(x .\left(\frac{y}{2}\right) .\left(\frac{y}{2}\right) .\left(\frac{z}{3}\right) .\left(\frac{z}{3}\right) .\left(\frac{z}{3}\right)\right)^{\frac{1}{6}}\leq \frac{1}{2}$
$\Rightarrow \large\left(x .\left(\frac{y}{2}\right) .\left(\frac{y}{2}\right) .\left(\frac{z}{3}\right) .\left(\frac{z}{3}\right) .\left(\frac{z}{3}\right)\right)\leq \frac{1}{64}$
$\Rightarrow x . y^{2} . z^{3} \leq \frac {108}{64} =\frac {27}{16}.$
Hence D) is the correct answer